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2 years ago

Which type of carbon fixation stores carbon dioxide in acid form? a. c3 b. c4 c. cam d. all of the above

Chemistry

2 answers:

Luba_88 [7]2 years ago

8 0

The type of carbon fixation stores carbon dioxide in acid form is CAM i.e. crassulacean acid metabolism.

CAM stands for crassulacean acid metabolism in this process photosynsthesis is occured at day time but the exchange of gases takes place at night itself only.

In this carbon fixation process, carbon dioxide is stored in the form of organic acid malic acid and losses carbon dioxide at the night time and by doing this it helps in the storage of water.

Hence option (C) is correct.

To know more about CAM, visit the below link:

brainly.com/question/4170802

Lyrx [107]2 years ago

3 0

Answer:

CAM

Explanation:

Yes the correct answer is option C, ( CAM) I just took the test.

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Ganezh [65]

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Questionnaires are the tool used for the research and the data collection for the analysis, investigation, and conclusion. It is one of the data collection tools that comprises questions and prompts.

Questionnaires are the part of surveys that gathers information on the topic of the research and is easy to analyze and cost-effective. It is a valuable tool that collects data from a large number of participants.

Therefore, the correct blank is the <u>questionnaire</u>.

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2 years ago

How many grams of CaCl2 are in 250 mL of 2.0 M CaCl2?

Tom [10]

The answer is:  " 56 g CaCl₂ " .
__________________________________________________________

Explanation:
__________________________________________________________
2.0 M CaCl₂ = 2.0 mol CaCl₂ / L ;

Since: "M" = "Molarity" (measurement of concentration);

= moles of solute per L {"Liter"} of solution.
__________________________________________________________
Note the exact conversion: 1000 mL = 1 L .

Given: 250 mL ;

250 mL = ? L ? ;

250 mL * (1 L / 1000 L) = (250/1000) L = 0.25 L .
___________________________________________________________

(2.0 mol CaCl₂ / L ) * (0.25L) = (2.0) * (0.25) mol  = 0.50 mol CaCl₂ ;

We have: 0.50 mol CaCl₂ ; Convert to "g" (grams):

→ 0.50 mol CaCl₂ .
___________________________________________________________
1 mol CaCl₂ = ? g ?

From the Periodic Table of Elements:

1 mol Ca = 40.08 g

1 mol Cl = <span>35.45 g .
</span>
There are 2 atoms of Cl in " CaCl₂ " ;

→ Note the subscript, "2", in the " Cl₂ " ;
__________________________________________________________
So, to calculate the molar mass of "CaCl₂" :

40.08 g + 2(35.45 g) =

40.08 g + 70.90 g = 110.98 g ; round to 4 significant figures;

→ round to 111 g/mol .
__________________________________________________________
So:

→ 0.50 mol CaCl₂ = ? g CaCl₂ ? ;

→ 0.50 mol CaCl₂ * (111 g CaCl₂ / mol CaCl₂) ;

= (0.50) * (111 g) CaCl₂ ;

= 55.5 g CaCl₂ ;

  → round to 2 significant figures;

  → 56 g CaCl₂ .
___________________________________________________________
The answer is: " 56 g CaCl₂ " .
___________________________________________________________

6 0

3 years ago

Read 2 more answers

What types of bonds or interactions are overcome when a nonmetal extended network melts?

STatiana [176]

Covalent bonds or interactions are overcome when a nonmetal extended network melts.

Typically, nonmetals form covalent bonds with one another. A polyatomic ion's atoms are joined by a form of link called covalent bonding. A covalent bond requires two electrons, one from each of the two atoms that are connecting.

One technique to depict the formation of covalent connections between atoms is with Lewis dot formations. The number of unpaired electrons and the number of bonds that can be formed by each element are typically identical. Each element needs to share an unpaired electron in order to establish a covalent bond.

Therefore, covalent bonds or interactions are overcome when a nonmetal extended network melts.

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4 0

2 years ago

Fission and Fusion of Atomic Nuclei:

bogdanovich [222]

Answer:

C B A D

Explanation:

6 0

3 years ago

0.53g of acetanilide was subjected to kjeldahl determination and the ammonia produced was collected in 50cm3 of 0.50M of h2so4.o

Lady bird [3.3K]

Answer:

10.57% of N in acetanilide

Explanation:

All nitrogen in the sample is converted in NH₃ in the Kjeldahl determination. The NH₃ reacts with H₂SO₄ as follows:

2NH₃ + H₂SO₄ → 2NH₄⁺ + SO₄²⁻

The acid in excess in titrated with Na₂CO₃ as follows:

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

To solve this question we must find the moles of sodium carbonate = Moles of H₂SO₄ in excess. The added moles - Moles in excess = Moles of sulfuric acid that reacts:

<em>Moles Na₂CO₃ anf Moles H₂SO₄ in excess:</em>

0.025L * (0.05mol / L) = 1.25x10⁻³ moles Na₂CO₃ / 0.01360L =

0.09191M * 0.250L = 0.0230 moles H₂SO₄ in excess.

<em>Moles H₂SO₄ added:</em>

0.050L * (0.50mol / L) = 0.0250 moles H₂SO₄ added

<em>Moles that react:</em>

0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄

<em>Moles of NH₃ = Moles N:</em>

0.0020 moles H₂SO₄ * (2mol NH₃ / 1mol H₂SO₄) = 0.0040 moles NH₃ = Moles N

<em>mass N and mass percent:</em>

0.0040 moles N * (14g / mol) = 0.056gN / 0.53g * 100 =

<h3>10.57% of N in acetanilide</h3>

7 0

2 years ago